3, 8, 24, 28, Eureka!

The news has been so unrelentingly bad these past few weeks that I’m taking momentary refuge in good old numerology. I happened to re-read this blog post by John Baez about the free modular lattice on 3 generators. This is a nice bit of pure math that features rather prominently the numbers 3, 8, 24 and 28. The numerological part is that I noticed the same numbers popping up in a problem that I had studied for other reasons, so I figured it would be fun to write about, even if my 28 isn’t exactly equal to Baez’s 28, so to speak.

Note that of these three matrices, only $\sigma_y$ is antisymmetric, and also note that we have
$$\sigma_z \sigma_x = -\sigma_x\sigma_z = i\sigma_y.$$

This much is familiar, though that minus sign gets around. For example, it is the fuel that makes the GHZ thought-experiment go, because it means that
$$\sigma_x \otimes \sigma_x \otimes \sigma_x = -(\sigma_x \otimes \sigma_z \otimes \sigma_z)(\sigma_z \otimes \sigma_x \otimes \sigma_z)(\sigma_z \otimes \sigma_z \otimes \sigma_x).$$

And this leads us to where the 8 comes into the story. Let’s consider the finite-dimensional Hilbert space made by composing three qubits. This state space is eight-dimensional, and we build the three-qubit Pauli group by taking tensor products of the Pauli matrices, considering the $2 \times 2$ identity matrix to be the zeroth Pauli operator. There are 64 matrices in the three-qubit Pauli group, and we can label them by six bits. The notation
$$\left(\begin{array}{ccc} m_1 & m_3 & m_5 \\ m_2 & m_4 & m_6\end{array}\right)$$
means to take the tensor product
$$(-i)^{m_1m_2} \sigma_x^{m_1} \sigma_z^{m_2} \otimes (-i)^{m_3m_4} \sigma_x^{m_3} \sigma_z^{m_4} \otimes (-i)^{m_5m_6} \sigma_x^{m_5} \sigma_z^{m_6}.$$

Now, we ask: Of these 64 matrices, how many are symmetric and how many are antisymmetric? We can only get antisymmetry from $\sigma_y$, and (speaking heuristically) if we include too much antisymmetry, it will cancel out. More carefully put: We need an odd number of factors of $\sigma_y$ in the tensor product to have the result be an antisymmetric matrix. Otherwise, it will come out symmetric. Consider the case where the first factor in the triple tensor product is $\sigma_y$. Then we have $(4-1)^2 = 9$ possibilities for the other two slots. The same holds true if we put the $\sigma_y$ in the second or the third position. Finally, $\sigma_y \otimes \sigma_y \otimes \sigma_y$ is antisymmetric, meaning that we have
$$9 \cdot 3 + 1 = 28$$
antisymmetric matrices in the three-qubit Pauli group. In the notation established above, they are the elements for which
$$m_1 m_2 + m_3 m_4 + m_5 m_6 = 1 \mod 2.$$

Puzzle: This has a secret geometrical meaning in terms of the Fano plane. What is it?

Hint: $28 = 7 \cdot 4 = 7 \cdot (7 – 3)$.

We have a 3 (the nontrivial elements of the single-qubit Pauli group), an 8 (the dimension of the three-qubit Hilbert space), and a 28 (the number of antisymmetric three-qubit Pauli operators). Does this story have a 24 as well? Sneakily, yes—because the same combinatorial notation that we used to enumerate the 28 antisymmetric Pauli operators also enumerates the 28 bitangents to a quartic curve. This is a lovely piece of nineteenth-century geometry, in the genre of the 27 lines on a cubic surface. It connects to Galois theory, as this paper explains:

The story of the free modular lattice on 3 generators is a story about how 3 things together build up an interesting collection of 28 things that live in an 8-dimensional space. I find it rather cute that an 8-dimensional space also yields an interesting collection of 28 things built up from 3 things in this other way.